 Boards are visible, it's okay? Okay, not immediately, okay? So first of all, I'd like to thank the organizers for inviting me here. It's great to be back in ICTP after several years. So my lectures will be on de-instanton effects in string theory. So let me write this. Is this writing visible from back? And I should also say that please ask questions whenever you have any instead of waiting till the end. So let me begin by just saying a few words about string theory, okay? And then I'll come to the instantons. So the original formulation of string theory was perturbative. So original formulation is perturbative, okay? Which means that we have an algorithm to generate a perturbation expansion of various quantities. So in particular, if you are given some background and this background is given in the form of a 2D CFT or you can think of this also as a compactification. So what we can compute is the following. That first, the spectrum of states which typically contain a few massless fields and infinite tower of massive fields, massive states, few massless and infinite tower of massive states. And the other thing we can compute is scattering amplitudes involving these states. But when I say that we can compute this, we can compute this only in a, as a perturbation series, okay? So let's take the example of case of scattering amplitudes, okay? So this typically has the form. So the scattering amplitudes in string theory are expressed as a sum. So this Gs is what is called the string coupling and the expansion is in parts of Gs. This is some constant, okay? Some integer typically or sometimes mostly it's integer in perturbation theory, okay? And these are the non-trivial object. These contain information about the scattering amplitudes. These are functions of the quantum number of the state, of the external states which includes momentum, for example. And string theory tells you how to calculate this, okay? At least it gives you an algorithm for how to calculate this, okay? And the algorithm involves integrals over modularized spaces of Riemann surfaces. Now if you haven't seen this before, you don't worry too much about it, right? Because you will not use this explicitly, okay? We'll encounter some integrals of this kind, but those will be reasonably simple kind, okay? And I'll explain when they come along, okay? But this is a general framework of string theory that we have a way to calculate scattering amplitudes as a perturbation expansion in the coupling, okay? And the coefficients of expansion are computable according to fixed algorithm. And that algorithm involves, you have to do integrals, certain integrals over modularized spaces of Riemann surfaces. Now from quantum field theory, we know that there is more to it than perturbation expansion, okay? So in QFT, we have additional contributions, non-parturbative contributions, okay? So these are the terms which cannot be, which basically whose Taylor series expansion vanishes to all orders, okay? So this is why they are not captured in perturbation expansion. And we expect that similar effects also exist in string theory, similar terms to be present in string theory. In quantum field theory, we know that such terms are present, okay? Because you have a path integral expression, okay? And manipulating that path integral, of course, you can generate the perturbation expansion through Feynman rules. But you also know that the path integral gets additional contribution from other saddle points, okay? And those are the instant terms, okay? So while we expect that similar contribution should be present in string theory, okay? We don't, a priori, have a systematic way to compute this, okay? Because we don't have a path integral formulation, okay? So we need a non-partramatic formulation like path integral systematically compute this. Now, unfortunately, at present, we don't have a full non-partramatic formulation. So at present, we don't have any one. There are some exceptions, okay? But those are very special, okay? In the sense that those are for a very specific background. And let me just list them anywhere. Exception, specific background, okay? For example, for some string background, we have a dual matrix model description, okay? Which is a full non-partramatic, well, you can try to make the matrix model into a non-partramatic description of string theory, okay? Although given a matrix model, it doesn't necessarily imply that you have a full non-partramatic formulation. ADHCFT certainly is an example where you can try to use dual conformal field theories to give non-partramatic formulation, okay? And most famous example is typed to be string theory on ADS5 times S5, which is given by a dual Young Mills theory. Or in some cases, even though you don't have the full non-partramatic control, we can use S duality to get some insight into the non-partramatic effects. But these apply for very specific backgrounds, okay? For a generic background, we don't know how to define string theory non-partramatically. No question? Yes. Is it known this AN, how do they behave for large N? Typically, we expect them to be an asymptotic series. Right? So they are not borderline summable. We don't expect them to be borderline summable. They are poles on the real axis. So that's the general expectation. Is it an expectation or is it a calculator ball in some instances? Well, see what people do calculate is the growth of the volume of the motorized space, right? Now that's just one aspect, right? I mean, we are not, and the full amplitude involves also correlation functions. And those are harder to calculate, right? So they estimate that one makes is just how, how fast the volume of the motorized space grows. Right? And it's based on that, we expect this to be not borderline summable. Okay, so these are some of the shortcomings of the current formulation of string theory. Nevertheless, it turns out that there is one class of effects, one class of nonpartisanal effects that we can compute in a wide class of backgrounds. So despite the shortcomings, there is one class of nonpartisanal effects that can be studied systematically. Yeah, and these are the re-instantal effects. Okay, and I'll describe what these are. Okay? So these give additional contribution to the amplitudes besides the perturbative contribution. So these gives additional contributions. And the general form of these additional contribution is, let me write down the general form first. So first, okay, let me first label this. So this is a constant. This beta is also a constant. And these BNs are, these are functions of the quantum numbers of external states, just like the ANs. And these are also given, just like the ANs, these are also given by integrals over modular space of Riemann surfaces. But this time, the Riemann surfaces also have bias boundaries. So first, you see that because of the C to the minus C over GS, if we try to do a Taylor series expansion of this around GS equal to zero, every term in the Taylor series expansion will vanish. Because if you take derivatives of C to the minus C over GS, however, many derivatives you take and evaluate it at GS equal to zero, the result is just zero, okay? If you go from the positive side. So this is the reason why you don't see this. This is distinct from the perturbation expansion. The second difference is that the power of GS, if beta is a fixed constant for a given amplitude, the power of GS, you see, there it is coming as parts of GS square, here it's the power of GS. This is some of the, these are some of the references that happen between the perturbative contribution and the non-parturbative contribution. But the main, yes? GS square. For D in sentence, it is this, right? With only one unique coefficient. With one? With just one coefficient, I mean, there cannot be any other coefficients like two C, three C. Yeah, because there may be different kinds of D in sentence or multi D in sentence, right? But it's always some constant, right? It's not that there is only single contribution of this kind, right? There are other contributions of C prime, right? Or two C, as you said. So those are certainly there. For a given D in sentence, this is the kind of contribution one gets. So the main problem, and this is what distinguishes the instant and contribution from parturbative contribution. They simply look like infinite integrals. And I'll just write down some examples. So some examples we encountered. So we'll see in our discussion later that you have an integral like this. This integral is finite at T, at T equal to zero because as T goes to zero, this vanishes. But clearly it's divergent from T equals to infinity, right? This diverges exponentially, this diverges logarithmically. Another example is, okay? This integral, you can see diverges from the y equal to zero n, right? This is one over a T y over y square, or this gives you T y over y, right? Both are divergent. Here is a more complicated example. A double integral. The omega are just constants, okay? The omega one, omega two, omega these are just constants. But what I want to draw your attention to is that these integrals are also divergent, both from the v equal to zero n, you can see that they are divergent integrals and from the x equal to zero n, okay? So when I said that we have algorithm to compute bn's, we have algorithm to write bn's as integrals, okay? And these integrals come from integrals of what the modular space of Riemann surfaces. But these integrals are divergent and the main issue is how to deal with these divergences, okay? So this will be the main goal of these lectures, is to extract unambiguous finite answers from them, okay? So let me write down the answers that we'll get eventually, right? But this is just, I'm writing down the answers right now and I mean, during the course of the lectures, I'll explain how these answers come. This integral, for example, we'll see we'll give you i over four pi square. This will be minus half and this will be minus half omega square log four, okay? So these will be seen in the lectures, okay? I'm just writing down these answers because we, so that we know what we are aiming for, right? We really want to get finite answers out of these integrals because these are the integrals that enter the expressions for these bn's, okay? And unless we can get finite answers out of this, we cannot claim that we have understood the n-sentence. Are there questions? The n-dependence? Yes, so these are for low values of n, right? As you go to higher and higher values of n, you get more and more complicated integrals. So n-dependence, I mean, this is for specific n that I have written down, okay? Because higher n will mean, there will be multi-dimensional integrals. The number of integrals and variables themselves will become larger, okay? And the integral and the divergences will also become more complicated. Any other questions? Now we'll also be able to test this procedure, although I may not discuss it in these lectures, okay? But we can't test this. So test, okay? We'll use it whenever a dual description exists, okay? Namely that I, we use the procedure to get a result like this. But then I say that in some cases, you also have dual description, okay? Which is a nonpart of the description that allows you to calculate these coefficients, okay? So if you do have a dual description, then it better be that whatever result you get by using this algorithm to extract finite answer sort of these, agree with what you get from the dual description, okay? So this is the test that we can perform, yes? So are these dual descriptions always unambiguous or do you also sometimes sort out an ambiguity in a dual description? Yeah, sometimes you sort out an ambiguity in the dual description, but sometimes it's also unambiguous, right? For example, the ace duality in type to be, right? Gives quite unambiguous prediction of what the instanton correction should be and we can check against that, right? Matrix model sometimes have ambiguities, right? And you can sort out those ambiguities using both sides. Right, so in other words, the formalism that you are going to describe has no ambiguities whatsoever. That's right, yeah. This has no ambiguity whatsoever. Okay, thanks. Well, the only ambiguity in this formalism is what you have in a quantum field theory that we have to, this ambiguity exists even in path integral. That when you write a path integral, you give the action and you have to specify an integration contour, right? That integration contour you choose based on various physical inputs, right? But if you change the integration contour, you may get a different result for the path integral, right? So that ambiguity is always there and that will also be present in this case. But that will be the only ambiguity that we'll have. You said that this is one class of non-perturbary effects and it's associated to some specific backgrounds. So which kind of backgrounds we are talking about and why this class is special with respect to other non-perturbary effects? Okay, so the backgrounds that we are talking about are type two string compactivations, type two are bosonic. I mean, where the world sheet is left-right symmetric, right? For example, the heterotic string theory doesn't have the incentives, okay? Or any d-brains, for example. But when you do have non-perturbary effects, these are the leading non-perturbary effects. Because the other non-perturbary effects, the usual instantons have to be minus c over gs squared, which are subdominant compared to this. So when the instantons are present, they are the dominant non-perturbary corrections. So it's very important that the n appears with the coefficient one. What appears with? The n in the particularly expansion, right? This one, no, not this, okay. This is related to this, but the important point is that the power of gs here is e to the minus c over gs and not c over gs squared. The usual instantons have e to the minus c over gs squared. So that's why they are more suppressed when the gs goes to zero limit. Okay, now before I go on, let me ask one question that is often asked in this context, okay? And I suspect that this question has also been asked already and that is the following. That does it make sense compute instanton corrections? Because as you saw, the perturbative series has this structure, sum over n gs to the power 2n plus alpha. And clearly every term here in this expansion is in the small gs limit is larger than this one. Okay, so I may ask is it, does it make sense to talk about instanton corrections before we even sum the perturbation series? So here I'll try to give two answers which one is practical but one is more conceptual. So the practical answer is that for many quantities, practical answer that for many quantities. So in such cases clearly it makes sense to talk about d instanton corrections because the perturbative contribution is completely understood, right? It either is not there at all, okay? So this happens for example in the computation of certain terms in the superpotential, right? So certain kind of terms, if you take in n equal to one super symmetric compactification in four dimensions, okay? Then certain class of terms are simply not generated in perturbation theory and they are the d instantons often are the leading contribution to such terms, right? So there it certainly makes sense to talk about d instanton contribution. In type 2b theory in 10 dimensions, okay? There are perturbative contribution to certain amplitudes but those series terminate, right? Other after one order up to second order, all hard terms vanish, okay? So then the perturbative contribution is completely understood and then the d instantons are the next leading corrections, okay? So in such cases clearly there is justification of why you want to study d instanton. There are also examples without super symmetry by the way, there are, we'll study this c equal to one Bosonic string theory in some detail, okay? And there we'll see that the certainness amplitudes have imaginary part and the imaginary part doesn't receive any contribution from the perturbative series, right? So they are d instantons are the leading contribution to the imaginary part of the amplitude. So they are examples of this kind. So this is a practical answer but there is also a conceptual answer and that is the following. So in QFT for example, instantons, QFT includes quantum mechanics. Instantons are subdominant saddles in the path integral, okay? They are subdominant in the sense that their contribution is exponential suppress compared to the leading contribution which is what the perturbative contribution is. And the instanton amplitudes, contribution to the amplitude from the steepest descent contour of the path, steepest descent contour of the saddle. So in the field space, if we think of the location of the various saddle points, okay? So in the field space, there may be various saddle points. This is the perturbative, this is the instanton one, this is the instanton two. And then for each of these saddles, there is a steepest descent contour, right? Then in higher dimension, the generalization of steepest descent contour is what are called left side symbols, okay? But there are some surfaces, some subspaces, okay? Along which in the, in this is the complex field space, okay? You always draw the steepest descent contour in the complex field space. So there are some subspaces that pass through these saddles. And when you talk about the instanton contribution to the amplitude, okay? What it is computing in perturbation theory is a result of the integral on those surfaces, those steepest descent contours. So it is true that if the steepest descent contour of the perturbative saddle gives much larger contribution compared to the steepest descent contour of the instanton contour, okay? Nevertheless, it certainly makes sense to ask, okay? What is the result of the steepest descent contour through the instanton, right? In the field space, if I specify a contour, I can ask what is the result of that integral, right? And that's what the instanton contribution computes. So it's an interesting quantity by itself, okay? Even though the actual integration contour may be involving both perturbative and some of the instanton saddles, and then this one may dominate, okay? But if you have chosen the integration contour to be the steepest descent contour of the instanton, then instanton amplitudes are all that you'll get. Now in string theory, of course you don't have a path integral description, a fundamental path integral description, but one expects that there should be a similar interpretation also in string theory. So when you say that you are trying to compute the instanton amplitudes in string theory, okay? Effectively, you are computing the integrated, result of path integral, okay? What about that maybe? The result of path integral over certain contour, right? Which passes through these non-perturbative saddles, okay? And the fact that they are dominated by the perturbative contour, okay? We do not need to take into account, right? This is a quantity that is of interest by itself. What is the result of the integral over the steepest descent contour of a given saddle? So that's a more conceptual reason why the instanton amp contribution to the amplitudes are useful quantities to study, okay? Even if in practice, they may give the subdomain and contribution. In quantum field theory, we also have some other non-perturbative contributions which has no such interpretation like a renormal ones. Okay? How do you know that these instanton says no, these are not of this kind? Well, I think we know it because of this fact that you have this e to the minus c over gs application, right? And this is typically the signal that there is a, there is a saddle whose action is c over gs, okay? And in fact, we can specifically identify the saddles as d brands, okay? So in that sense, I mean, you already know that there exist saddles, right? Even though we don't have a path integral description, we, in strength, we can identify solutions to equations of motion, okay? Just by saying that they are conformal field theories, right? So in that sense, you already know that there are saddles with action of this kind, okay? So it's more natural to interpret the instanton contribution as coming from the steepest descent contour of the saddles. Okay, so let me then say what the instantons are. And here I assume that you have some idea of what d brands are, okay? Because d instantons are special kind of d brands. So d instantons, okay? And I'll use what d instanton is a somewhat general essence. So we'll define d instantons as a follow, as a following. These are d brands with Dirichlet boundary condition, all non-compact directions, including time, okay? And we normally describe d instantons in the Euclidean spacetime. So in the Euclidean time, okay? So in order to describe a d brand, you have to specify for each spacetime direction, whether you impose Dirichlet or Neumann boundary condition, okay? Sometimes you may also consider mixed boundary condition, but let's just do Dirichlet and Neumann. So for each spacetime direction, you have to specify whether you're considering Dirichlet or Neumann boundary condition, okay? Here the idea is that along all the non-compact directions, you have to put Dirichlet boundary condition. Along the compact spacetime directions, you may put Neumann or Dirichlet, okay? All of those will be classified as d instantons, different kinds of d instantons. But because you are putting the Dirichlet boundary condition on the non-compact direction, this means that these are localized all non-compact directions, okay? In particular, it's localized at some particular value of time. And because these are localized along non-compact directions, what they describe as finite action solutions and strain theory, okay? And the action is, has a form, some constant over gs, okay? And this follows from the standard argument that all d-branes, okay? The tension of a d-brane, it goes as one over g string, some constant over g string. At the same argument, it goes as so, tension here goes over to action, because it's also localized in the time direction. And the standard calculation that you can find in Paul Chinsky's book can be used to calculate the disc coefficients, okay? So the action of the d-instanton, you can compute. Given, once you specify what boundary condition you're imposing, on which direction, you can explicitly compute the action of the d-instanton following the standard technique. So it is in this sense that d-instantons are generally saddles of strain theory, right? Even though we don't know what path integral describes in strain computation. So these are analogous to instantons in quantum field theory. And except that in normal instantons in quantum field theory we'll have a g-square, right? The action typically goes as one over g-square, here it goes as one over g, right? That's the main difference. And this give nonpartner very contribution to string amplitudes, okay? These amplitudes, as I have already written, takes the form to the minus c over gs and dn gs to the power n plus beta. And these are what are given by integrals over the modular spaces of human surfaces with boundaries. Why with boundaries? Because once you have d-blend, right? You have open strings which end on d-blends, right? And the open string wall sheets have boundaries, okay? That's the reason why the d-instanton amplitudes have boundaries, the wall sheets have boundaries. Now, in these lectures we'll consider single d-instantons, but that's just for simplicity, okay? The procedure that I'll be describing can also be generalized to multiple d-instantons, okay? It just requires a little more work. For multiple instantons, suppose there are n-instantons, the first difference will be that there'll be e to the minus n c over gs, okay? Because there are n-instantons total action is n c over gs. And then the calculation of bn's will be a little more complicated, right? Because the boundary can now have different boundary conditions because it can end on the first d-instanton or the second d-instanton or the third d-instanton and so on. So now, let me give some systematics of d-instanton amplitudes, okay? Because this will be important. So first, we now have, we have open and closed strings in the presence of d-instantons. So if you think of d-instanton as a point in space time, okay, at least among the non-compact directions, it may have extension in the compact directions, which I have not drawn. Then there can certainly be open strings whose ends are on the d-instanton, okay? So these are open strings. But we can also have closed strings in the vicinity of the d-instantons, right? Or far away from the d-instantons. So both of those kinds of strings are present. But there is a crucial difference between the open strings and the closed strings, okay? Namely the open strings, okay? These exist only for limited time. Because d-instantons are localized in time, right? They occur at some particular value of time. And because the ends of the open string are tied to the d-instanton, right? The open string cannot extend the wall volume of the open string or the wall sheet of the open string cannot extend too much in the time direction, right? It has to be confined in the vicinity of this where the d-instanton is located. So because of this, okay? These cannot be asymptotic states. These are not asymptotic states. So the interpretation of the open strings is that the d-instantons are localized. Yes, go ahead. Sorry, just a quick clarification conceptually. So of course the instanton is perfectly localized in time. So when you say limited time, you mean because they also will have a finite energy excitation. So you can allow a vicinity also in time. Or do you actually mean that it just exists for one instant in time? Well, d-instantons exist for one instant of time. But the open string, of course, the middle of the open string can extend in time, right? It cannot go very far away from the time, right? They will cause infinite energy, energy, or... Yeah, that was my question. So it is related to finite energy. That's right, yeah. Okay. That if you stretch it too much in time, right? That's like infinite tension. So open strings are not asymptotic states. So when we talk about d-instanton amplitudes, okay? The external states, whose scattering amplitudes we calculate, okay? Are always closed strings, okay? It's like a normal instanton in quantum field theory, right? If we take instantons in quantum field theory, what do they compute? They compute the usual scattering amplitude of the excitation of the quantum field theory, right? The instantons have their own modes, right? The instanton can get translated, it can vibrate, it can do various things, okay? But those are not the modes whose scattering amplitudes we calculate, okay? Those modes are useful in computing scattering amplitudes of the external states, which are the usual states of quantum field theory, okay? So here, they are in the same spirit. Open strings are describing the dynamics of the d-instanton, okay? But they are not asymptotic states. So when we talk about scattering amplitudes, the external states will always be closed strings. This is something we should keep in mind. Is it okay? Are there questions on this? The second important point, okay? Which is again different from the usual perturbation theory, is that because we are imposing this boundary condition, okay? We have d-instanton located at a particular time, for example, right? So time transition invariance is broken. It's located at some particular value of the space coordinate. So space transition invariance is broken. And because this transition invariance is broken, that means that there is a energy momentum conservation by the usual wall sheets, okay? Is there wall sheets with boundaries? Because if the boundary is which break the transition invariance, okay? Eventually, of course, we'll see how the transition invariance is restored, how the energy momentum conservation is restored, okay? But at the level of computation, okay? When we use the standard string calculation, okay? Transition invariance is broken, and hence we don't have energy momentum conservation. And this has a non-trivial consequence, okay? Namely that now disconnected wall sheets, okay? Are on the same footing as connected wall sheets. You think in terms of quantum field theory Feynman diagrams, right? When you draw Feynman diagrams in quantum field theory, we are both disconnected and connected diagrams, right? In QFT, if you are doing four-point scattering, you can draw four-point scattering diagrams like this, okay? You can have four-point scattering diagrams like this. This is disconnected, this is connected. Normally, the reason that you don't worry about these is because these diagrams require additional energy momentum conservation than the overall energy momentum conservation, okay? This one, for example, will tell us that the energy momentum of this should be equal to the energy momentum of this, and energy momentum of this should be equal to energy momentum, okay? Which means there are two separate energy momentum conservation, okay? That have to be satisfied for this diagram. Whereas for this diagram, there is only one overall energy momentum conservation. So if we take a generic external momentum, right? Which satisfies just that overall energy momentum conservation. Then typically for generic external momentum, this diagram will not contribute because the delta function constant will not be satisfied. Otherwise, in quantum field theory, these diagrams are there, and similarly in string theory also, disconnected wall sheets are very much present. But because deans and tons don't have this feature that individual wall sheets satisfy energy momentum conservation, okay? They are in no way different from the standard wall sheets which are connected, right? So you have to include all of them. In, for the instanton amplitudes, we need to include disconnected wall sheets, yes. When you say that translation invariance is broken, you mean spontaneously broken, right? The time transition and also the space that transition are going on compact directions. Yes, okay. I mean from the point of view of wall sheet, it is, you are choosing a different volume in which the translation invariance is spontaneously broken, is it correct? Well, I would not say the spontaneously broken, it's just, there's a boundary, right? And at the boundary, the energy is not conserved, right? It can just flow out of the boundary, or the momentum is not conserved. That's what is happening, right? Because at the boundary, you don't, if the boundary is the one you're just breaking translation invariance. That's, I would say it's more explicitly broken because you have just broken it by putting this boundary condition on the wall sheet. But keeping into account that the instantons are dynamical object in the full theory, translation invariance will be restored. Yes, yes, it will be restored eventually, right? But at the level of individual wall sheets, they will not be restored. We'll see later how the full translation invariance is restored, but individual wall sheets don't have that property. So now, let me describe how to organize the instanton amplitudes, okay? Because you want to do this power series, right? Which are the ones which give dominant contribution in this power series, okay? And here we are going to use the usual rules of string perturbation theory, okay? And that is the following that wall sheet with Euler number chi gives contribution of order gs to the minus chi, okay? That's the standard rule in string perturbation theory and that holds even for, in the presence of the instantons. So this means that in order to get the highest negative power of gs, that means lowest 10 in this expansion, we have to maximize the Euler number. And this is, okay? So how do you maximize the Euler number? Okay? Discs have Euler number one. These sphere, of course, have more, but sphere has no boundaries. So we are looking for surface with boundaries, right? Because those are there, we can also explore the fact that the energy momentum is not concept. Anulus, okay? Anulus is this topologically, right? A disc with a hole. This has Euler number zero, okay? And then you can list various other things, but these are the ones which are low Euler number. High Euler number, okay? Others, there are also many surface with negative Euler number. So because of this fact that the discs of Euler number one, we see that the way to get the maximum inverse power of gs is to maximize the number of discs. So to get the leading contribution, we need to maximize the number of discs. So this is the first observation. Second observation is that because Anulus has Euler number zero, in respect to how many Anulus you use, the power of gs remains the same, right? Because it has gs to the zero. So we can use, can use as many Anulus as gone, without a cost in gs. There's a question? Yes. Should I think of the Anulus as one instant or as two instantons? No, one instanton only because when I say Anulus, on both boundaries, we'll put the same boundary condition. So the instanton is just one point in space tab, right? The number of strings that can end on it can be anything. So Anulus at both ends of the open string are ending on the instanton, right? So it's like, you can think of this as a loop, take this one open string, and it's a loop of that open string. So the instanton action is still one over gs. It's not two over gs. No, it's not two over gs, it's still one over gs. Right? So given a single instanton, you can have as many wordsheet as you want ending on that. There are just different excitations of the same instanton. So the leading contribution, okay? With this we can write down the leading contribution, okay? So the leading contribution has the following structure. We have this exponential of minus c over gs, which is the suppressant due to the action, okay? Then we have the exponential of the Anulus, okay? Because you can use as many as you want. So you can use zero or one or two, et cetera. And the combinatoric factor is such that it just exponentiates. And then we have a product of disk one-point functions. So these are the closed string vertex operators. So you have n closed strings, right? The idea is that you take n disks, insert one closed string on each of these disks. So you cannot have disk without a vertex operator because that vanishes, right? Because of this, there is a sl2 or volume you have to divide by. And that makes a disk without vertex operator vanish. So you have to insert at least one vertex operator. So you maximize the number of disks by inserting each external vertex operator on one disk. Maybe you could say a couple more words why this exponentiates exactly. If you have an arbitrary number of boundaries but you tie them up to annular. I mean, you said the combinatorics works out but it's maybe not obvious that it does. Yeah, okay. Basically, okay. So first of all, one corresponds to no annulus, in this expansion. Then you can have one annulus. For the two annulus, I don't know how that I can do it often. Basically, there is a one over two factorial because the exchange of these two annulus gives the same diagram, right? That's the argument, right? And you can repeat this for any annulus, right? But I think if you want more details of this I have to work it out. Don't want to do it right now on this spot. Okay, but roughly that's the idea that you have multiple annulus that is the exchange symmetry, right? That gives you one over n factorial and that's why it actually exponents. So this exponential basically comes from these all of these. Yeah, it's roughly that exactly, yeah. It's like a zero point function in quantum field theory. Hi, I should. So are there a fermionic zero modes that you have to soak up? I mean, do you have to? Yeah, yeah. So if there are fermionic zero modes you have to do a little more work, right? So this is a leading contribution provided the answer is non-zero, right? I mean, it may happen that in some cases this answer is zero, right? Then you have to look for what is the first non-zero contribution. So we'll discuss the fermion zero modes but certainly if there are fermionic zero modes you have to insert them appropriately here. So next order, for the leading order there's nothing else, right? This is all you have. At the next order what you can do, okay? There are many ways you can increase power of gs, right? You want to increase the power of gs by one. So what we can do is we can give up one disk. So instead of having each, that each disk having one closed string vortex of water maybe one disk has two and the rest have one. So here is a possible contribution at next order. Okay, so this basically means that we have one less disk. So we lose the factor of gs, right? Because on one of the disks we insert two more closed string vortex of water so we are using a disk. So this is one contribution but there are others at the same order, okay? You can have this. You can replace a disk by annulus. So all of them are disks except that one disk is replaced by annulus compared to this. This is over here, right? So again you lose one over gs because this instead of being one over gs is a water one. Or you can do, you have these as before but then you add to this a disk with two holes, okay? Disk with two holes have Euler character minus one. This has chi equal to minus one, okay? So because of this we lose the factor of gs, right? It's gs to the minus k chi so you just get a factor of gs. So these are the various ways you can build the other question. Now the problem comes because most of these diagrams are divergent, okay? Disk with one disk one point function is okay. Okay, that's in fact the only one which is finite. But this one for example is infinite. We'll write down there as a result for this next time for some theories, right? This is infinite, okay? And the question is what do you do about this? These are both infinite. These two point function annulus one point function are all infinite, okay? They and some of the integrals I showed you essentially come from this. So we have to understand what the origin of these infinities are and then understand how to extract finite result of these infinities, right? So that will be the goal of this next two lectures, okay? I think my time is up so I'll stop here. We can have some questions or comments or.